How can I prove $$\|AB\|_p ≤ \|A\|_p \|B\|_p$$ where $$\|A\|_p = \max_{x≠0}\frac{\|Ax\|_p}{\|x\|_p}, \qquad p = 1, 2, 3, \ldots$$ and known information below:

(i) $\|A\| ≥ 0$, with equality iff $A = 0.$

(ii) $\|cA\| = |c| \|A\|$, for any $c ∈ \mathbb R.$

(iii) $\|A + B\| ≤ \|A\| + \|B\|$.

*Not duplicate of Proof of matrix norm property: submultiplicativity, I don't know whats the connection between that answer an my question.

  • $\begingroup$ I have saw that, but I don't understand that, I mean, it's not p-norm in that answer. $\endgroup$ – hcnak Jul 23 '17 at 6:16
  • $\begingroup$ I don't see connection between that answer and my question. @MartinR, the answer in that question may use some concept I don't know yet. $\endgroup$ – hcnak Jul 23 '17 at 6:18
  • $\begingroup$ It holds for any matrix norm. Just write $||ABx||_p \le ||A||_p ||Bx||_p \le ||A||_p ||B||_p ||x||_p$ as in the other answer. $\endgroup$ – Martin R Jul 23 '17 at 6:18
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    $\begingroup$ It is a duplicate. The linked question is about a general induced norm, of which the $p$-norm is a special case. $\endgroup$ – user1551 Jul 23 '17 at 7:29